Consider the series $f_n(x) = \sum_{1}^{n}\frac{nx^2}{n^3+x^3}$ I'm trying to prove that this series does not uniformly converge on $[0,\infty)$
Let $$\lim_{n \to \infty} \sum_{1}^{n}\frac{nx^2}{n^3+x^3} = F(x)$$
While disproving the book has said :
$\lim_{x \to \infty} F(x) = \infty$ while $\lim_{x \to \infty} \frac{nx^2}{n^3+x^3} = 0$ so it can not uniformly converge. My question is how did it deduce that $\lim_{x \to \infty}F(x)= \infty$ I couldn't get that?
It is easy to verify that $a_k=\frac{kx^2}{k^3+x^3}$ is nonnegative and decreasing sequence, so the integral test tells us:
\begin{align} \int_1^\infty \frac{tx^2\mathrm d t}{t^3+x^3}\leq \sum\limits_{k=1}^{\infty}\frac{kx^2}{k^3+x^3} =F(x) \end{align} Note that \begin{align} \int_1^\infty \frac{tx^2\mathrm d t}{(t+x)^3} \leq\int_1^\infty \frac{tx^2\mathrm d t}{t^3+x^3} \end{align} And integration by parts yields: \begin{align} \int_1^\infty \frac{tx^2\mathrm d t}{(t+x)^3} =\frac{x^3+2x^2} {2(1+x)^2} \end{align} The inequality becomes: \begin{align} \frac{x^3+2x^2} {2(1+x)^2}\leq\sum\limits_{k=1}^{\infty}\frac{kx^2}{k^3+x^3} =F(x) \end{align}
It holds for all $x\in[0, \infty) $, so \begin{align} \lim\limits_{x \to \infty}\frac{x^3+2x^2} {2(1+x)^2}\leq\lim\limits_{x \to \infty} F(x) \end{align} Since $\lim\limits_{x \to \infty}\frac{x^3+2x^2} {2(1+x)^2}=\infty$, we have $\lim\limits_{x \to \infty} F(x) =\infty$.